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0

$begingroup$

Let $a$ be a sequence of the squares for all natural numbers.

So: $a = 0^2, 1^2, 2^2, 3^2, 4^2,5^2,... = 0, 1, 4, 9,16,25,...$

Is there a way to find out when the equation below will be true without checking each of the terms?

$x + a_i = a_j$

where $x,i,j in mathbbN $

For example:

Let $x = 23$, then:
$23 + 121 = 144$

so $i=11$ and $j=12$

Edit: To clarify I would like to calculate $i$ and $j$ from $x$.

sequences-and-series number-theory

share|cite|improve this question

edited Apr 2 at 17:05

Mateusz Michalak-Palarz

asked Mar 30 at 22:55

Mateusz Michalak-PalarzMateusz Michalak-Palarz

11

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Odd numbers can be made easily by $(n+1)^2-n^2=2n+1$
  $endgroup$
  – Sil
  Mar 30 at 23:02
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $x$ can be any natural number which is not $2$ more than a multiple of $4$
  $endgroup$
  – Henry
  Mar 30 at 23:12
  

  
  
  
  

add a comment | 

0

$begingroup$

Let $a$ be a sequence of the squares for all natural numbers.

So: $a = 0^2, 1^2, 2^2, 3^2, 4^2,5^2,... = 0, 1, 4, 9,16,25,...$

Is there a way to find out when the equation below will be true without checking each of the terms?

$x + a_i = a_j$

where $x,i,j in mathbbN $

For example:

Let $x = 23$, then:
$23 + 121 = 144$

so $i=11$ and $j=12$

Edit: To clarify I would like to calculate $i$ and $j$ from $x$.

sequences-and-series number-theory

share|cite|improve this question

edited Apr 2 at 17:05

Mateusz Michalak-Palarz

asked Mar 30 at 22:55

Mateusz Michalak-PalarzMateusz Michalak-Palarz

11

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Odd numbers can be made easily by $(n+1)^2-n^2=2n+1$
  $endgroup$
  – Sil
  Mar 30 at 23:02
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $x$ can be any natural number which is not $2$ more than a multiple of $4$
  $endgroup$
  – Henry
  Mar 30 at 23:12
  

  
  
  
  

add a comment | 

$begingroup$

Let $a$ be a sequence of the squares for all natural numbers.

So: $a = 0^2, 1^2, 2^2, 3^2, 4^2,5^2,... = 0, 1, 4, 9,16,25,...$

Is there a way to find out when the equation below will be true without checking each of the terms?

$x + a_i = a_j$

where $x,i,j in mathbbN $

For example:

Let $x = 23$, then:
$23 + 121 = 144$

so $i=11$ and $j=12$

Edit: To clarify I would like to calculate $i$ and $j$ from $x$.

sequences-and-series number-theory

share|cite|improve this question

edited Apr 2 at 17:05

Mateusz Michalak-Palarz

asked Mar 30 at 22:55

Mateusz Michalak-PalarzMateusz Michalak-Palarz

11

$endgroup$

share|cite|improve this question

edited Apr 2 at 17:05

Mateusz Michalak-Palarz

asked Mar 30 at 22:55

Mateusz Michalak-PalarzMateusz Michalak-Palarz

11

share|cite|improve this question

edited Apr 2 at 17:05

Mateusz Michalak-Palarz

asked Mar 30 at 22:55

Mateusz Michalak-PalarzMateusz Michalak-Palarz

11

asked Mar 30 at 22:55

Mateusz Michalak-PalarzMateusz Michalak-Palarz

11

asked Mar 30 at 22:55

Mateusz Michalak-PalarzMateusz Michalak-Palarz

11

2 Answers
2

active

oldest

votes

1

$begingroup$

For any $x$, consider all the factorisations of $x$ as $y times z$ where $y$ and $z$ have the same parity, i.e. both even or both odd, and $y ge z$; this will be possible if and only if $x$ is not $2$ more than a multiple of $4$

Then since $left(fracy+z2right)^2 - left(fracy-z2right)^2 = yz$ we have $x + left(fracy-z2right)^2 = left(fracy+z2right)^2$

In your example, with $x=23$ we only have $23times 1$ giving $23 + left(frac23-12right)^2 = left(frac23+12right)^2$

For another example, if $x=45$ then we have $45 times 1 = 15 times 3 = 9 times 5$ so we get:

• $45 + 22^2 = 23^2$


• $45 + 6^2 = 9^2$


• $45 + 2^2 = 7^2$

You want $i=left(fracy-z2right)^2$ and $j=left(fracy+z2right)^2$

share|cite|improve this answer

edited Mar 30 at 23:26

answered Mar 30 at 23:21

HenryHenry

101k482170

$endgroup$

add a comment | 

0

$begingroup$

You can look for a pattern, in fact, letting $i=n, j=n+1$ and $x=2n+1$ yields:
$$a_n+2n+1=a_n+1$$
This can easily be confirmed by letting $a_n=n^2$, indeed $n^2+2n+1=(n^2+1)^2$.

share|cite|improve this answer

answered Mar 30 at 23:24

Arsene1412Arsene1412

114

$endgroup$

add a comment | 

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1

$begingroup$

For any $x$, consider all the factorisations of $x$ as $y times z$ where $y$ and $z$ have the same parity, i.e. both even or both odd, and $y ge z$; this will be possible if and only if $x$ is not $2$ more than a multiple of $4$

Then since $left(fracy+z2right)^2 - left(fracy-z2right)^2 = yz$ we have $x + left(fracy-z2right)^2 = left(fracy+z2right)^2$

In your example, with $x=23$ we only have $23times 1$ giving $23 + left(frac23-12right)^2 = left(frac23+12right)^2$

For another example, if $x=45$ then we have $45 times 1 = 15 times 3 = 9 times 5$ so we get:

• $45 + 22^2 = 23^2$


• $45 + 6^2 = 9^2$


• $45 + 2^2 = 7^2$

You want $i=left(fracy-z2right)^2$ and $j=left(fracy+z2right)^2$

share|cite|improve this answer

edited Mar 30 at 23:26

answered Mar 30 at 23:21

HenryHenry

101k482170

$endgroup$

add a comment | 

1

$begingroup$

For any $x$, consider all the factorisations of $x$ as $y times z$ where $y$ and $z$ have the same parity, i.e. both even or both odd, and $y ge z$; this will be possible if and only if $x$ is not $2$ more than a multiple of $4$

Then since $left(fracy+z2right)^2 - left(fracy-z2right)^2 = yz$ we have $x + left(fracy-z2right)^2 = left(fracy+z2right)^2$

In your example, with $x=23$ we only have $23times 1$ giving $23 + left(frac23-12right)^2 = left(frac23+12right)^2$

For another example, if $x=45$ then we have $45 times 1 = 15 times 3 = 9 times 5$ so we get:

• $45 + 22^2 = 23^2$


• $45 + 6^2 = 9^2$


• $45 + 2^2 = 7^2$

You want $i=left(fracy-z2right)^2$ and $j=left(fracy+z2right)^2$

share|cite|improve this answer

edited Mar 30 at 23:26

answered Mar 30 at 23:21

HenryHenry

101k482170

$endgroup$

add a comment | 

$begingroup$

For any $x$, consider all the factorisations of $x$ as $y times z$ where $y$ and $z$ have the same parity, i.e. both even or both odd, and $y ge z$; this will be possible if and only if $x$ is not $2$ more than a multiple of $4$

Then since $left(fracy+z2right)^2 - left(fracy-z2right)^2 = yz$ we have $x + left(fracy-z2right)^2 = left(fracy+z2right)^2$

In your example, with $x=23$ we only have $23times 1$ giving $23 + left(frac23-12right)^2 = left(frac23+12right)^2$

For another example, if $x=45$ then we have $45 times 1 = 15 times 3 = 9 times 5$ so we get:

• $45 + 22^2 = 23^2$


• $45 + 6^2 = 9^2$


• $45 + 2^2 = 7^2$

You want $i=left(fracy-z2right)^2$ and $j=left(fracy+z2right)^2$

share|cite|improve this answer

edited Mar 30 at 23:26

answered Mar 30 at 23:21

HenryHenry

101k482170

$endgroup$

share|cite|improve this answer

edited Mar 30 at 23:26

answered Mar 30 at 23:21

HenryHenry

101k482170

answered Mar 30 at 23:21

HenryHenry

101k482170

answered Mar 30 at 23:21

HenryHenry

101k482170

0

$begingroup$

You can look for a pattern, in fact, letting $i=n, j=n+1$ and $x=2n+1$ yields:
$$a_n+2n+1=a_n+1$$
This can easily be confirmed by letting $a_n=n^2$, indeed $n^2+2n+1=(n^2+1)^2$.

share|cite|improve this answer

answered Mar 30 at 23:24

Arsene1412Arsene1412

114

$endgroup$

add a comment | 

0

$begingroup$

You can look for a pattern, in fact, letting $i=n, j=n+1$ and $x=2n+1$ yields:
$$a_n+2n+1=a_n+1$$
This can easily be confirmed by letting $a_n=n^2$, indeed $n^2+2n+1=(n^2+1)^2$.

share|cite|improve this answer

answered Mar 30 at 23:24

Arsene1412Arsene1412

114

$endgroup$

add a comment | 

$begingroup$

You can look for a pattern, in fact, letting $i=n, j=n+1$ and $x=2n+1$ yields:
$$a_n+2n+1=a_n+1$$
This can easily be confirmed by letting $a_n=n^2$, indeed $n^2+2n+1=(n^2+1)^2$.

share|cite|improve this answer

answered Mar 30 at 23:24

Arsene1412Arsene1412

114

$endgroup$

share|cite|improve this answer

answered Mar 30 at 23:24

Arsene1412Arsene1412

114

answered Mar 30 at 23:24

Arsene1412Arsene1412

114

answered Mar 30 at 23:24

Arsene1412Arsene1412

114